Problem: Let $g(x)=x^2\ln(x)$. What is the absolute minimum value of $g$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1}{e^2}$ (Choice B) B $\dfrac{\ln(2)}{e}$ (Choice C) C $-\dfrac{1}{2e}$ (Choice D) D $g$ has no minimum value
Let's first find the relative extremum points of $g$, and then consider them along with the function's end behavior in both directions. Note that the domain of $g$ is all real numbers such that $x>0$. We start with finding the critical points of $g$. The derivative of $g$ is $g'(x)=x(2\ln(x)+1)$. $g'(x)=0$ for $x=\dfrac{1}{\sqrt e}$. [What about x=0?] $g'$ is defined for all real numbers in the domain of $g$. Therefore, our only critical point is $x=\dfrac{1}{\sqrt e}$. Our critical point divides the function's domain (which is all real numbers such that $x>0$ ) into two intervals: $0$ $1$ $(0,\frac{1}{\sqrt e})$ $(\frac{1}{\sqrt e},\infty)$ $\frac{1}{\sqrt e}$ Let's evaluate $g'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g'(x)$ Verdict $\left(0,\dfrac{1}{\sqrt e}\right)$ $x=\dfrac12$ $g'\left(\dfrac12\right)=\dfrac12 - \ln(2)<0$ $g$ is decreasing $\searrow$ $\left(\dfrac{1}{\sqrt e},\infty\right)$ $x=1$ $g'(1)=1>0$ $g$ is increasing $\nearrow$ Let's imagine ourselves walking on the graph of $g$, starting at the left end of the domain $($ which is $0$ $)$ and going all the way to the right (until $+\infty$ ). According to the table, we will start by going down until we reach $x=\dfrac{1}{\sqrt e}$. Then, we will be forever going up. Therefore, $g$ must obtain its absolute minimum value at $x=\dfrac{1}{\sqrt e}$. We are asked to find that minimum value, which is $g\left(\dfrac{1}{\sqrt e}\right)=-\dfrac{1}{2e}$. In conclusion, the absolute minimum value of $g$ is $-\dfrac{1}{2e}$.